이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include "robots.h"
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int MAXN = 1 << 20;
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define fi first
#define se second
#define all(v) (v).begin(), (v).end()
#define fillchar(a, s) memset((a), (s), sizeof(a))
void compress (vector<int> &v) {
sort(all(v));
v.resize(unique(all(v)) - v.begin());
}
int A, B, N;
int *X, *Y, *W, *S;
int ordw[MAXN], ords[MAXN];
struct cmpw {
bool operator() (int x, int y) {
return pii(W[x], S[x]) < pii(W[y], S[y]);
}
};
struct cmps {
bool operator() (int x, int y) {
return pii(S[x], W[x]) < pii(S[y], W[y]);
}
};
bitset<MAXN> vis;
bool moo (int g) {
if (g * ll(A + B) < N) {
return false;
}
priority_queue<int, vector<int>, cmps> pqs;
priority_queue<int, vector<int>, cmpw> pqw;
//what is the limit here?
int rem = N;
vis.reset();
//memory limit is so fucking strict...i don't even understand
int ptr = 0;
for (int i = 0; i < A; i++) {
//insert stuff in
for (; ptr < N && W[ordw[ptr]] < X[i]; ptr++) {
pqs.push(ordw[ptr]);
}
for (int taken = 0; !pqs.empty() && taken < g; taken++) {
vis[pqs.top()] = true;
rem--;
pqs.pop();
}
}
ptr = 0;
for (int i = 0; i < B; i++) {
for (; ptr < N && S[ords[ptr]] < Y[i]; ptr++) {
int ind = ords[ptr];
if (!vis[ind]) {
pqw.push(ind);
}
}
for (int taken = 0; !pqw.empty() && taken < g; taken++) {
vis[pqw.top()] = true;
rem--;
pqw.pop();
}
}
return rem == 0;
}
int putaway (int aaa, int bbb, int nnn, int xxx[], int yyy[], int www[], int sss[]) {
A = aaa;
B = bbb;
N = nnn;
X = xxx;
Y = yyy;
W = www;
S = sss;
sort(X, X + A);
sort(Y, Y + B);
for (int i = 0; i < N; i++) {
if (W[i] >= X[A - 1] && S[i] >= Y[B - 1]) {
return -1;
}
ordw[i] = i;
ords[i] = i;
}
sort(ordw, ordw + N, cmpw());
sort(ords, ords + N, cmps());
//memory is the issue here
int lo = 0, hi = N;
while (hi - lo > 1) {
int mid = (lo + hi) / 2;
if (moo(mid)) {
hi = mid;
} else {
lo = mid;
}
}
return hi;
}
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